Integration is the inverse process of differentiation

Hi everyone,
I know that integration is the inverse process of differentiation, and that the definite integral is defined as:
[itex]\int_{a}^{b} f(x) dx = \lim_{n \to \infty} \sum^{n}_{i = 1} f(x_i) \Delta x[/itex]
assuming that the integrand is defined over the interval [a,b].

My question is: Why is it that the antiderivatives give the area under the curve? For example:

[itex]\int \frac{1}{1+x^2} dx = \arctan{x} + C[/itex]

How does arctan give the area under curve of the integrand between a and b?

An anti-derivative is not the area under the curve. An integral is. That being said, the area under the curve is the geometric interpretation. The best way to see it is to look at the definition of the integral itself. Sf(x)dx means sum of f(x) which is the y value, times dx which is the delta x segments on the x axis. This means, height (y value) * base (dx) which is the area of a rectangle. the S is the sum of all the rectangles. The fundamental theorem of calculus is the algebraic interpretation, the "getting back from the derivative".

The fundamental theorem of calculus:
If dF/dx= f(x), then [tex]\int f(x)dx= F(b)- F(a)

and
If [itex]F(x)= \int_a^x f(t)dt[/itex], then dF/dx= f(x)

One way to show that is this: If y= f(x)> 0, between x= a and x= b, then we can identify (or define) [tex]\int_a^x f(t)dt[/itex] as "the area of the region bounded by the graphs of y= f(x), y= 0, x= a, and x= b".

Given any x0, [itex]F(x_0)= \int_a^{x_0} f(t)dt[/itex] is the area under the curve from a to x0 and, for any h> 0, [itex]F(x+h)= \int_a^{x_0+ h}f(t)dt[/itex] is the area under the curve from a to x0+ h. The difference, F(x0+ h)- F(x0) is the area under the curve from x0 to x0+ h.

Suppose f(x*) is the minimum value for f on that interval, x0 to x0+ h. Then clearly the rectangle with base h and height f(x*) is completely contained in that region. Suppose f(x*) is the maximum value for f on that interval, then the rectangle with base h and height f(x*) contains that region. In terms of area, [itex]hf(x_*)\le \int_{x_0}^{x_0+ h} f(x) dx\le hf(x^*)[/itex]. Dividing by h,
[tex]f(x_*)\le \frac{\int_{x_0}^{x_0+ h} f(t)dt}{h}\le f(x^*)[/tex]
but the area between x0 and x0+ h is just the area from a to x0 minus the area from a to x0. That is:
[tex]f(x_*)\le \frac{F(x+h)- F(x_0)}{h}\le f(x^*)[/tex]

Now take the limit a h-> 0. Since x0+ h goes to x0, and both x* and x*[/sub] lie between them, both x* and x* go to x0 so
[tex]f(x_0)\le \frac{dF}{dx}(x_0)\le f(x_0)[/itex]
That is, dF/dx evaluated at x=x0, is just f(x0).